\(\int (a (b \sec (c+d x))^p)^n \, dx\) [69]

Optimal result

Integrand size = 14, antiderivative size = 83 \[ \int \left (a (b \sec (c+d x))^p\right )^n \, dx=-\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-n p),\frac {1}{2} (3-n p),\cos ^2(c+d x)\right ) \left (a (b \sec (c+d x))^p\right )^n \sin (c+d x)}{d (1-n p) \sqrt {\sin ^2(c+d x)}} \] Output:

-cos(d*x+c)*hypergeom([1/2, -1/2*n*p+1/2],[-1/2*n*p+3/2],cos(d*x+c)^2)*(a* 
(b*sec(d*x+c))^p)^n*sin(d*x+c)/d/(-n*p+1)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \left (a (b \sec (c+d x))^p\right )^n \, dx=\frac {\cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n p}{2},\frac {1}{2} (2+n p),\sec ^2(c+d x)\right ) \left (a (b \sec (c+d x))^p\right )^n \sqrt {-\tan ^2(c+d x)}}{d n p} \] Input:

Integrate[(a*(b*Sec[c + d*x])^p)^n,x]
 

Output:

(Cot[c + d*x]*Hypergeometric2F1[1/2, (n*p)/2, (2 + n*p)/2, Sec[c + d*x]^2] 
*(a*(b*Sec[c + d*x])^p)^n*Sqrt[-Tan[c + d*x]^2])/(d*n*p)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4611, 3042, 4259, 3042, 3122}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \left (a (b \sec (c+d x))^p\right )^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a (b \sec (c+d x))^p\right )^ndx\)

\(\Big \downarrow \) 4611

\(\displaystyle (b \sec (c+d x))^{-n p} \left (a (b \sec (c+d x))^p\right )^n \int (b \sec (c+d x))^{n p}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (b \sec (c+d x))^{-n p} \left (a (b \sec (c+d x))^p\right )^n \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n p}dx\)

\(\Big \downarrow \) 4259

\(\displaystyle \left (\frac {\cos (c+d x)}{b}\right )^{n p} \left (a (b \sec (c+d x))^p\right )^n \int \left (\frac {\cos (c+d x)}{b}\right )^{-n p}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (\frac {\cos (c+d x)}{b}\right )^{n p} \left (a (b \sec (c+d x))^p\right )^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{-n p}dx\)

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {\sin (c+d x) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-n p),\frac {1}{2} (3-n p),\cos ^2(c+d x)\right ) \left (a (b \sec (c+d x))^p\right )^n}{d (1-n p) \sqrt {\sin ^2(c+d x)}}\)

Input:

Int[(a*(b*Sec[c + d*x])^p)^n,x]
 

Output:

-((Cos[c + d*x]*Hypergeometric2F1[1/2, (1 - n*p)/2, (3 - n*p)/2, Cos[c + d 
*x]^2]*(a*(b*Sec[c + d*x])^p)^n*Sin[c + d*x])/(d*(1 - n*p)*Sqrt[Sin[c + d* 
x]^2]))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

rule 4611

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Simp[b^ 
IntPart[p]*((b*(c*Sec[e + f*x])^n)^FracPart[p]/(c*Sec[e + f*x])^(n*FracPart 
[p]))   Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p}, x] 
&&  !IntegerQ[p]
 

Maple [F]

\[\int \left (a \left (b \sec \left (d x +c \right )\right )^{p}\right )^{n}d x\]

Input:

int((a*(b*sec(d*x+c))^p)^n,x)
 

Output:

int((a*(b*sec(d*x+c))^p)^n,x)
 

Fricas [F]

\[ \int \left (a (b \sec (c+d x))^p\right )^n \, dx=\int { \left (\left (b \sec \left (d x + c\right )\right )^{p} a\right )^{n} \,d x } \] Input:

integrate((a*(b*sec(d*x+c))^p)^n,x, algorithm="fricas")
 

Output:

integral(((b*sec(d*x + c))^p*a)^n, x)
 

Sympy [F]

\[ \int \left (a (b \sec (c+d x))^p\right )^n \, dx=\int \left (a \left (b \sec {\left (c + d x \right )}\right )^{p}\right )^{n}\, dx \] Input:

integrate((a*(b*sec(d*x+c))**p)**n,x)
 

Output:

Integral((a*(b*sec(c + d*x))**p)**n, x)
 

Maxima [F]

\[ \int \left (a (b \sec (c+d x))^p\right )^n \, dx=\int { \left (\left (b \sec \left (d x + c\right )\right )^{p} a\right )^{n} \,d x } \] Input:

integrate((a*(b*sec(d*x+c))^p)^n,x, algorithm="maxima")
 

Output:

integrate(((b*sec(d*x + c))^p*a)^n, x)
 

Giac [F]

\[ \int \left (a (b \sec (c+d x))^p\right )^n \, dx=\int { \left (\left (b \sec \left (d x + c\right )\right )^{p} a\right )^{n} \,d x } \] Input:

integrate((a*(b*sec(d*x+c))^p)^n,x, algorithm="giac")
 

Output:

integrate(((b*sec(d*x + c))^p*a)^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a (b \sec (c+d x))^p\right )^n \, dx=\int {\left (a\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^p\right )}^n \,d x \] Input:

int((a*(b/cos(c + d*x))^p)^n,x)
 

Output:

int((a*(b/cos(c + d*x))^p)^n, x)
 

Reduce [F]

\[ \int \left (a (b \sec (c+d x))^p\right )^n \, dx=b^{n p} a^{n} \left (\int \sec \left (d x +c \right )^{n p}d x \right ) \] Input:

int((a*(b*sec(d*x+c))^p)^n,x)
 

Output:

b**(n*p)*a**n*int(sec(c + d*x)**(n*p),x)